JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 12)
Let $$\mathrm{S}=\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z+i|=|z-1|\}$$. Then the set of all values of $$x$$, for which $$w=2 x+i y \in \mathrm{S}$$ for some $$y \in \mathbb{R}$$, is :
$$\left(-\sqrt{2}, \frac{1}{2 \sqrt{2}}\right]$$
$$\left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right]$$
$$\left(-\sqrt{2}, \frac{1}{2}\right]$$
$$\left(-\frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}\right]$$
Explanation
$S:\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z-i|=|z-1|\}$
$$ |z-1+i| \geq|z| $$
_29th_July_Evening_Shift_en_12_1.png)
$|z|<2$
_29th_July_Evening_Shift_en_12_2.png)
$$ |z-i| = |z-1| $$
_29th_July_Evening_Shift_en_12_3.png)
$\because \quad w \in S$ and $w=2 x+i y$
$2 x<\frac{1}{2} \quad\quad \therefore x<\frac{1}{4}$
$(2 x)^{2}+(-2 x)^{2}<4$
$4 x^{2}+4 x^{2}<4$
$x^{2}<\frac{1}{2} \Rightarrow x \in\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\therefore \quad x \in\left(-\frac{1}{2}, \frac{1}{4}\right]$
$$ |z-1+i| \geq|z| $$
$|z|<2$
$$ |z-i| = |z-1| $$
$\because \quad w \in S$ and $w=2 x+i y$
$2 x<\frac{1}{2} \quad\quad \therefore x<\frac{1}{4}$
$(2 x)^{2}+(-2 x)^{2}<4$
$4 x^{2}+4 x^{2}<4$
$x^{2}<\frac{1}{2} \Rightarrow x \in\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\therefore \quad x \in\left(-\frac{1}{2}, \frac{1}{4}\right]$
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